Recurrence relations Worked Solutions — Further Maths Maths

Fully worked, step-by-step solutions to Further Maths Recurrence relations questions. See exactly how to solve problems on recurrence-relations, iteration, first-order, closed-form.

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Further Maths70 questionsStep-by-step solutions
Question 1
2 markseasy
A sequence is defined by un+1=3un+2u_{n+1}=3u_{n}+2, n≥1n\ge 1, with u1=4u_{1}=4. Find the value of u4u_{4}.

Worked solution

  1. Apply the recurrence with n=1n=1

    u2=3⋅4+(2)=14u_{2}=3\cdot 4+\left(2\right)=14

    Each new term is built from the term immediately before it.

  2. Apply the recurrence with n=3n=3

    u4=3⋅44+(2)=134u_{4}=3\cdot 44+\left(2\right)=134

    Each new term is built from the term immediately before it.

  3. State the required term

    u4=134u_{4}=134

    This is the value asked for.

Answer
134134
Question 2
2 markseasy
A sequence is defined by un+1=2un−5u_{n+1}=2u_{n}-5, n≥0n\ge 0, with u0=3u_{0}=3. Find the value of u5u_{5}.

Worked solution

  1. Apply the recurrence with n=0n=0

    u1=2⋅3+(−5)=1u_{1}=2\cdot 3+\left(-5\right)=1

    Each new term is built from the term immediately before it.

  2. Apply the recurrence with n=1n=1

    u2=2⋅1+(−5)=−3u_{2}=2\cdot 1+\left(-5\right)=-3

    Each new term is built from the term immediately before it.

  3. Apply the recurrence with n=4n=4

    u5=2−27+(−5)=−59u_{5}=2-27+\left(-5\right)=-59

    Each new term is built from the term immediately before it.

  4. State the required term

    u5=−59u_{5}=-59

    This is the value asked for.

Answer
−59-59
Question 3
2 markseasy
A sequence is defined by un+1=4unu_{n+1}=4u_{n}, n≥0n\ge 0, with u0=3u_{0}=3. Find an expression for unu_{n} in terms of nn.

Worked solution

  1. Solve the homogeneous recurrence

    un+1=4un ⇒ un=A4nu_{n+1}=4u_{n}\ \Rightarrow\ u_{n}=A4^{n}

    Ignoring f(n)f(n) leaves a geometric sequence with common ratio 44, so the complementary function is A4nA4^{n}.

  2. Use the initial condition u0=3u_{0}=3

    A=3 ⇒ A=3A = 3\ \Rightarrow\ A=3

    Substituting n=0n=0 into the general solution gives one linear equation, and a first-order recurrence has only one constant to find.

  3. State the closed form

    un=3⋅4nu_{n}=3\cdot 4^{n}

    This expression reproduces every term of the sequence directly from nn.

Answer
3⋅4n3\cdot 4^{n}
Question 4
2 markseasy
A sequence is defined by un+1=2unu_{n+1}=2u_{n}, n≥1n\ge 1, with u1=6u_{1}=6. Find an expression for unu_{n} in terms of nn.

Worked solution

  1. Solve the homogeneous recurrence

    un+1=2un ⇒ un=A2nu_{n+1}=2u_{n}\ \Rightarrow\ u_{n}=A2^{n}

    Ignoring f(n)f(n) leaves a geometric sequence with common ratio 22, so the complementary function is A2nA2^{n}.

  2. Use the initial condition u1=6u_{1}=6

    2A=6 ⇒ A=32 A = 6\ \Rightarrow\ A=3

    Substituting n=1n=1 into the general solution gives one linear equation, and a first-order recurrence has only one constant to find.

  3. Check the closed form against u2u_{2}

    u2=12from the recurrence, and the closed form gives 12u_{2}=12\quad\text{from the recurrence, and the closed form gives }12

    The closed form reproduces the term generated by the recurrence, which confirms it.

  4. State the closed form

    un=3⋅2nu_{n}=3\cdot 2^{n}

    This expression reproduces every term of the sequence directly from nn.

Answer
3⋅2n3\cdot 2^{n}
Question 5
2 markseasy
A sequence is defined by un+1=un+5u_{n+1}=u_{n}+5, n≥0n\ge 0, with u0=2u_{0}=2. Find an expression for unu_{n} in terms of nn.

Worked solution

  1. Recognise an arithmetic sequence

    un+1−un=5u_{n+1}-u_{n}=5

    Consecutive terms differ by the same constant, so the sequence is arithmetic with common difference 55.

  2. Write down the general term of an arithmetic sequence

    un=u0+5nu_{n}=u_{0}+5n

    Starting from u0u_{0}, the constant 55 is added once for every step taken.

  3. State the closed form

    un=2+5nu_{n}=2+5n

    This expression reproduces every term of the sequence directly from nn.

Answer
2+5n2+5n

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